Answer:
The element oxidized is the one whose oxidation number INCREASES......
Explanation:
And thus......
+
I
I
2CuO
+
C
→
2
0
C
u
+
C
O
2
↑
⏐
⏐
And here while copper (as
C
u
2
+
) is reduced, carbon is oxidized,
0
C
→
+
I
V
C
.
The second reaction represents the reduction
C
r
(
V
I
+
)
to
C
r
(
I
I
I
+
)
C
r
2
O
2
−
7
+
14
H
+
+
6
e
−
→
2
C
r
3
+
+
7
H
2
O
(
i
i
)
And the oxidation of ethanol to carbon dioxide:
H
3
C
−
C
H
2
O
H
+
H
2
O
→
2
C
O
2
+
8
H
+
+
8
e
−
(
i
i
i
)
i.e.
H
3
−
I
I
I
C
−
−
I
C
H
2
O
H
→
2
+
I
V
C
O
2
We takes
4
×
(
i
i
)
+
3
×
(
i
i
i
)
to eliminate the electrons.....
3
H
3
CC
H
2
O
H
+
3
H
2
O
+
4
C
r
2
O
2
−
7
+
56
H
+
+
24
e
−
→
8
C
r
3
+
+
28
H
2
O
+
6
C
O
2
+
24
H
+
+
24
e
−
To give finally............
3
H
3
CC
H
2
O
H
+
4
C
r
2
O
2
−
7
+
32
H
+
→
8
C
r
3
+
+
25
H
2
O
+
6
C
O
2
Dichromate ion
is reduced to
chromic ion
